How do you divide #(4+2i )/( 1-i)#?
You must eliminate the complex number in the denominator by multiplying by its conjugate:
1 + 3i
Require the denominator to be real . To achieve this multiply the numerator and the denominator by the complex conjugate of the denominator.
If (a + bi ) is a complex number then (a - bi) is the conjugate
here the conjugate of (1 - i ) is (1 + i )
# ((4 + 2i )(1 + i ))/((1 - i)(1 + i )) #
distribute the brackets to obtain :
# (4 + 6i + 2i^2 )/(1 - i^2 ) #
# i^2 = (sqrt(-1)^2 )= - 1 #
# (4 + 6i - 2 )/(1 + 1 ) = (2 + 6i )/2 = 2/2 + (6i)/2 = 1 + 3i#